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Non-conformal Unitary Mapping

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to find a unitary operator that can map two (two-dimensional) pure states [itex] |+\rangle, |-\rangle [/itex] as follows:
    [tex] |+\rangle \to \cos\theta |+\rangle + \sin\theta |-\rangle [/tex]
    [tex] |-\rangle \to \sin\theta |+\rangle + \cos\theta |- \rangle [/tex]
    For an arbitrary angle [itex] 0 \leq \theta \leq \frac\pi4 [/itex]

    3. The attempt at a solution

    The first obvious attempt at a solution is to simply create a linear system of equations for an element of [itex] U(2) [/itex], and solve, which gives
    [tex] \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} [/tex]
    However, this is obviously not unitary. Since I know that unitary mappings are not forced to be conformal, I think the mapping does exist, but am unsure where to go from here.
     
  2. jcsd
  3. Sep 16, 2009 #2

    nrqed

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    Are you sure the second term with a cos theta should not be negative? That's what is usually used, so that the kets in the new basis are still orthonormal (assuming that [tex] |+ \rangle [/tex] and [tex] |- \rangle [/tex] form an orthonormal basis)
     
  4. Sep 16, 2009 #3
    I had originally thought that myself, and so double checked before posting. According to the text I am using (and the associated diagram), the problem statement is correct.
     
  5. Sep 16, 2009 #4

    Avodyne

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    It's not possible. The original states are orthogonal, and the transformed states are not, but unitary transformations preserve inner products.
     
  6. Sep 16, 2009 #5
    I never said that the original two states were orthogonal. I realize that by using the +/- notation you may have assumed as such (as I realize they are commonly used to represent an orthonormal basis), but they needn't be for the purpose of this discussion.
     
  7. Sep 16, 2009 #6
    I'm not sure if it helps at all, but the value of theta isn't entirely arbitrary. I said so originally to see if it could be done in general. For my purpose, theta is

    [tex] \theta = \frac12 \arcsin(\langle + | - \rangle ) [/tex]

    That is,

    [tex] \sin2\theta = \langle + | - \rangle [/tex]
     
  8. Sep 16, 2009 #7

    nrqed

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    Ah ok! But you assumed too that they were orthonormal when you wrote the transformation matrix. You assumed that [tex] |+ \rangle [/tex] was the column vector with entries (1,0) and you assumed that [tex] |- \rangle [/tex] was the column vector with entries (0,1). You need to write the general expressions for those two states.

    By the way, what does "conformal" have to do with the question?
     
  9. Sep 16, 2009 #8
    Very good point, I did end up assuming they were orthogonal myself. I mentioned the non-conformality in that most unitary operators seem to be rotations, though I think that for this purpose we are primarily interested in a non-conformal map. I could be wrong.
     
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